Rationalize the denominator: cubic root of 2y^4/6x^4

- anonymous

Rationalize the denominator: cubic root of 2y^4/6x^4

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- anonymous

\[\sqrt[3]{2y ^{4}/6x ^{4}}\]

- anonymous

My instinct is to go down instead of making it \[x ^{64}\] because that would look dumb, unless it's correct but I've never had to decrease... do I just multiply by a negative exponent?

- anonymous

=cube root (2y^4)/cube root(6x^4)
=cube root(2y^4)*cuberoot(6x^4)/cube root(6x^4)*cube root(6x^4)
=cube root(2y^4)*cuberoot(6x^4)/6x^4
basically separate top and bottom roots
multiply by conjugates
fix ugly maths

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## More answers

- jhonyy9

cuberoot(y/3x)^4 =(y/3x)cuberoot(y/3x)=(ycuberoot3xy)/3x

- jhonyy9

2/6 simplifie by 2 and get 1/3

- jhonyy9

ok ?

- anonymous

Hmm, well that is handy, didn't know oyu could simplify it but what would the correct conjugate be?

- anonymous

you*

- jhonyy9

good luck
bye

- jhonyy9

1/sqrt3 =(sqrt3)/3 ---like example

- jhonyy9

ok ?

- anonymous

No actually, I'm confused... can you just tell me what I need to multiply by to get rid of the radical in the denominator?

- jhonyy9

1/sqrt3 so when there is radical in denominator multiplie numerator and denominator by conjugate so in this case by sqrt3 and in this way you get in denominator 3 and in numerator sqrt3

- jhonyy9

because sqrt3 * sqrt3 =3

- anonymous

I know... but what would the conjugate be. I know how to do all this already i get it I just need to know if the conjugate I am using is correct...

- jhonyy9

your answer not is right

- jhonyy9

check my answer step by step

- razor99

we have to simplify that.

- anonymous

Ok look, the denominator is cubic root of 6x^4 .... so what do I multiply it by to take a perfect cubic root?

- anonymous

sorry that is wrong

- anonymous

\[\sqrt[3]{36x^2}\]

- saifoo.khan

If sat is here, then there was no need of calling me! :)

- anonymous

then you will get in the denominator a perfect cube, namely
\[\sqrt[3]{216x^6}\]

- anonymous

He just got here thanks anyway

- anonymous

you got this?

- anonymous

you have
\[\sqrt[3]{6x^4}\] so to make it a perfect cube, you need the exponent on the x to be 6, and you have to multiply by 6^2=36 to make the constnant a cube

- anonymous

Wait why does the exponent for x have to be 6?

- anonymous

because you have a cube root so you need a number divisible by 3

- anonymous

i mean an exponent divisible by 3

- anonymous

\[\sqrt[3]{216x^6}=6x^2\]

- anonymous

So that would come out as \[6x ^{2}\] ?

- anonymous

yes

- anonymous

Ah thank you that is ALL I needed, I guess nobody understood what I meant. Thank you very much

- anonymous

yw

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